# Trouble understanding internal and external forces [duplicate]

systems can be defined and redefined and so the meaning of what an internal or an external force is also changes: I think im wrong here: if then we consider planet earth a system, all the things ( cars moving etc ) that happen in this system, are they internal forces,?

if there is a change in momenta inside the system, then it is caused by a force. no doubt. internal forces. why then is it said that an internal force cant cause a change in momentum?

what i think: the system's momentum as a whole doesnt change .. but i cant go anywhere from that fact for some reason. im stuck

## marked as duplicate by sammy gerbil, glS, Aaron Stevens, Jon Custer, user259412Nov 11 '18 at 19:23

• Yes. Definition of internal vs. external forces is purely subjective. A force can be either depending on what you take your system to be. – Aaron Stevens Nov 7 '18 at 10:32
• Related question by OP physics.stackexchange.com/questions/439419/… – Aaron Stevens Nov 7 '18 at 11:28
• You need to be very careful about how you define your system. Then, you need to answer your questions with respect to that particular system definition. – David White Nov 9 '18 at 20:56

You are not wrong. You seem to have understood it correctly. You just aren't fully accepting it yet.

If planet Earth is considered a system, then any force between crashing cars, bouncing balls and landing skydivers is internal.

Many of such force include momentum changes. Drop a rock and the ground slows it down to zero speed by having its momentum transferred. But remember that the Earth likewise absorbs this momentum. Seen from the outside, the system hasn't gained any - a part of the system lost some while another part gained some momentum. In total, the change is zero for the system, while it can be non-zero for the individual parts of the system.

Of this reason: Internal force cannot cause momentum change! If it does cause momentum change, then it must have been an external force. Which is why we in many cases can ignore internal forces.

If you consider the whole universe with everything in it as one huge system - or if you have a system which is isolated (no external forces acting), then you now have the usual momentum conservation law: the change in momentum is always zero; non is ever lost or gained. If momentum is lost or gained, then you must have made a mistake by not having the system isolated.

$$\let \mr=\mathrm \def\10#1#2{#1\cdot10^{{#2}}}\def\qy#1#2{#1\,\mr{#2}}$$ I think there is something more to be said, which might be useful.

1) Deciding "your" system.

It's true: to decide which is the system we are going to consider is widely at our choice. But it's not entirely arbitrary. For instance, if you are interested in studying a double star it would be senseless to neglect one of them. On the other side, if some object is very weakly influenced by other objects, whereas these are strongly influenced by the former, it makes sense to take the former as a fixed environment and concentrate ourselves on the latter, assuming them as subjected to external forces.

If you are asking yourself how this may be the case, given Newton's third law, think e.g. of a very important example: Earth and motions of a multitude of bodies on its surface or around it (with one exception: the Moon).

Newton' third law states equality of forces, but equal forces can have very different effects on two bodies, in inverse relation to their masses. Some numbers may help. Earth mass is about $$\qy{\10 6{24}}{kg}$$. If a stone block of mass $$\qy{10^4}{kg}$$ falls to ground from an height of $$\qy{10}m$$ it acquires a speed of $$\qy{14}{m/s}$$. Earth too is accelerated and "falls" towards the stone. Its final speed in c.o.m. frame will be $$\qy{\10{2.3}{-20}}{m/s}$$. For any foreseeable future, this Earth's motion is undetectable.

The same holds for every mechanical phenomenon on and around Earth: our planet may be viewed as a fixed environment, acting via gravitational force on all other bodies, satellites included. Moon is to be excepted both because of its mass (1/80 of Earth's) and of observable effects on Earth's ground - tides.

An analogous argument holds for solar system, where Sun's mass is dominant wrt planets' masses (let alone asteroids and comets). But here we must be careful, since mass ratios are not so big. Between Sun and Earth ratio is 330,000 but between Sun and Jupiter it's only about 1000. If you consider that astronomical measurements are extremely precise (8 significant digits are common in several cases) you see at once that Sun's motion may not be neglected, and solar system has to be studied as a whole, as a unique mechanical system, unless you content yourself with low-precision results.

2) Changing "your" system.

From all that I wrote up to now you can see that system's choice is not forever. It depends on problem, on precision you require, on quantities you are interested in. A large part of a physicist's work is in deciding how to deal with the problem he/she is given.

BTW, a much related matter is how to choose the reference frame. It too is not forced by the objects you are studying, but strongly depends on same things I said about defining your system. In many cases a frame integral with Earth is OK and you may also treat it as an inertial frame. In other cases non-inertial effects are important. And so on.

All this is not to discourage you - on the contrary. As I see it, a good part of the beauty of physics is in this continuous effort to find the best conceptual frame for the problem at hand. As you go on studying physics you will find it presenting itself all the time.